Photons Have Momentum — Question 4

Question 4

Compton energy loss and recoil angle follow the same vector balance

A photon of energy $E=m_ec^2=511\,\mathrm{keV}$ scatters through $90^\circ$ from a free electron initially at rest. For scattering from an initially resting free electron, use $E\prime=E/[1+(E/m_ec^2)(1-\cos\theta)]$, $\lambda\prime-\lambda=\lambda_C(1-\cos\theta)$, and $\lambda_C=h/(m_ec)$.

1. Find the scattered photon energy and electron kinetic energy.

2. Find the recoil-electron angle below the original photon direction.

Question 4 - Solution

Compton energy loss and recoil angle follow the same vector balance

(a) Method and calculation

$$\boxed{E\prime=E/(1+1)=255.5\,\mathrm{keV}}.$$$$\boxed{K_e=E-E\prime=255.5\,\mathrm{keV}}.$$

(b) Interpretation and further calculation

$$p_{ex}c=E-E\prime\cos\theta=E,\quad |p_{ey}|c=E\prime\sin\theta=E/2.$$$$\boxed{\tan\varphi=|p_{ey}|/p_{ex}=1/2},\quad\boxed{\varphi=26.5651^\circ}.$$

Physical check

The electron recoils on the opposite side of the incident axis from the scattered photon, preserving transverse momentum.